Q74 / 160 · AP EAMCET 2024 (22 May Shift 1) · Mathematics

If $x_1, x_2, x_3 \ldots x_n$ are $n$ observations such that $\Sigma\left(x_i+2\right)^2=28 n$ and $\Sigma\left(x_i-2\right)^2=12 n$, then the variance is:

  • Correct answer
    A
    12
  • B
    14
  • C
    16
  • D
    20

Solution

$\begin{aligned} & \text { Given that } \sum\left(x_i+2\right)^2=28 n \\ & \Rightarrow \sum x_i^2+4 \sum x_i+4 n=28 n \\ & \Rightarrow \sum x_i^2+4 \sum x_i=24 n ....(i)\end{aligned}$ Similarly $\sum\left(x_i-2\right)^2=12 n \Rightarrow \sum x_i^2-4 \sum x_i=8 n$
On solving equation (i) and (ii), we get $\sum x_i^2=16 n \text { and } \sum x_i=2 n$ $\therefore$ Variance $=\frac{\sum x_i^2}{n}-\left(\frac{\sum x_i}{n}\right)^2$ $=\frac{16 n}{n}-\left(\frac{2 n}{n}\right)^2=16-4=12$

Hence, the correct answer is option: 'A'.

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